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## Main Question or Discussion Point

Sorry about twin paradox again, I'll try to keep this simple as simple as I can.

Initial situation:

All participants are at rest, A1 and A2 are together at midpoint between B1 and B2. All clocks are in sync.

_____________A1, A2

_____B1____________________B2

----

A1 and A2 move, say, with speed 0,8c to opposite directions, measured in B's frame (the relative speed between A's being somewhat higher of course).

____________<-A1 A2->

_____B1____________________B2

---

When A2 and B2 meet, A2 makes instantaneous turnaround and begins to move to A1's direction with the same speed, thus entering A1's reference frame.

___<-A1__________________<-A2

_____B1____________________B2

When A's compare their clocks, they note that A2's clock is behind since A2 is the one that has made the acceleration. (1)

A's are inertial observers from now on, so this clock difference should be permanent.

---

B1 and B2 catch up the respective A's with identical acceleration, B1 and B2 never moving relative to each other. All four are now in the same reference frame. This is the end situation.

___<-A1__________________<-A2

___<-B1__________________<-B2

---

Different observers cannot disagree of local events, so the clock readings at meeting points should be absolute. Let

tA1 = clock time of A1 when A1 and B1 meet

tA2 = clock time of A2 when A2 and B2 meet

tB1 = clock time of B1 when A1 and B1 meet

tB2 = clock time of B2 when A2 and B2 meet

The events "A1 and B1 meet" and "A2 and B2 meet" are simultaneous in the B's frame.

Just before the B's accelerate, B's agree that

tA1 = tA2, tB1 = tB2 and tA1 < tB1. (2)

B's do not move relative to each other, so after acceleration is still tB1 = tB2. Because A1 and B1 are at the same place during the acceleration, their clocks go at the same rate; as with A2 and B2. Therefore, after the acceleration, B's still agree of (2).

(1) and (2) seem to contradict, all four observers nevertheless being in the same reference frame.

Does this have something to do with the "identical" acceleration of B's, or is there a more embarassing flaw somewhere? I'm not actually believing this is a contradiction ;)

Initial situation:

All participants are at rest, A1 and A2 are together at midpoint between B1 and B2. All clocks are in sync.

_____________A1, A2

_____B1____________________B2

----

A1 and A2 move, say, with speed 0,8c to opposite directions, measured in B's frame (the relative speed between A's being somewhat higher of course).

____________<-A1 A2->

_____B1____________________B2

---

When A2 and B2 meet, A2 makes instantaneous turnaround and begins to move to A1's direction with the same speed, thus entering A1's reference frame.

___<-A1__________________<-A2

_____B1____________________B2

When A's compare their clocks, they note that A2's clock is behind since A2 is the one that has made the acceleration. (1)

A's are inertial observers from now on, so this clock difference should be permanent.

---

B1 and B2 catch up the respective A's with identical acceleration, B1 and B2 never moving relative to each other. All four are now in the same reference frame. This is the end situation.

___<-A1__________________<-A2

___<-B1__________________<-B2

---

Different observers cannot disagree of local events, so the clock readings at meeting points should be absolute. Let

tA1 = clock time of A1 when A1 and B1 meet

tA2 = clock time of A2 when A2 and B2 meet

tB1 = clock time of B1 when A1 and B1 meet

tB2 = clock time of B2 when A2 and B2 meet

The events "A1 and B1 meet" and "A2 and B2 meet" are simultaneous in the B's frame.

Just before the B's accelerate, B's agree that

tA1 = tA2, tB1 = tB2 and tA1 < tB1. (2)

B's do not move relative to each other, so after acceleration is still tB1 = tB2. Because A1 and B1 are at the same place during the acceleration, their clocks go at the same rate; as with A2 and B2. Therefore, after the acceleration, B's still agree of (2).

(1) and (2) seem to contradict, all four observers nevertheless being in the same reference frame.

Does this have something to do with the "identical" acceleration of B's, or is there a more embarassing flaw somewhere? I'm not actually believing this is a contradiction ;)